Let $(W_t)_{t \geq 0}$ be a Brownian motion. Consider $$ X_t = \int^t_0 \text{sign}(W_u) \, dW_u $$ where $$\text{sign}(x) := \begin{cases} 1, & x \geq 0, \\ -1, & x>0. \end{cases}$$ Prove that $X_t$ and $W_t$ are independent.

How do I prove that? Naturally I need to compute $E[X_tW_t]=0$. I don't know if it helps me but I have computed $X^2_t-t$ which is a martingale so by Levy characterization Theorem $X_t$ is a brownian motion itself.

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    $\begingroup$ $E(X_tW_t)=E\left(\int_0^t\operatorname{sgn}(W_s)dW_s\int_0^tdW_s\right)=E\int_0^t\operatorname{sgn}(W_s)ds=0$. But I'm not sure if $X_t$ and $W_t$ are independent. $\endgroup$ – AddSup Jan 22 at 12:14
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    $\begingroup$ As a consequence of Tanaka's formula, $\int_0^t\text{sign}(W_s)\,dW_s\le |W_t|$ almost surely. This would seem to preclude the independence of these random variables. $\endgroup$ – John Dawkins Jan 22 at 15:03

It might be tempting to reason as follows: Since $(X_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ are Brownian motions, we know that $W_t$ and $X_t$ are Gaussian. On the other hand, it is possible to show (see @Addsup's comment) that $\mathbb{E}(X_t W_t) = 0$, and consequently $X_t$ and $W_t$ are uncorrelated. Since uncorrelated Gaussian random variables are independent, it follows that $X_t$ and $W_t$ are independent.

The reasoning is wrong. Why? In order to conclude that $W_t$ and $X_t$ are independent, we need to know that the random vector $(X_t,W_t)$ is Gaussian; it is not enough to know that (the marginals) $X_t$ and $W_t$ are Gaussian.

In fact, $W_t$ and $X_t$ are not independent. By Itô's formula, we have $$W_t^2 = 2 \int_0^t W_s \,dW_s + t.$$ As $\mathbb{E}(X_t)=0$ we thus find

\begin{align*} \mathbb{E}(W_t^2 X_t)& =2 \mathbb{E} \left( X_t \int_0^t W_s \, dW_s \right) \\ &=2 \mathbb{E} \left(\left[ \int_0^t \text{sgn}(W_s) \, dW_s \right] \left[ \int_0^t W_s \, dW_s \right] \right) \end{align*}

Applying Itô's isometry we obtain that

$$\mathbb{E}(W_t^2 X_t) =2 \mathbb{E} \left( \int_0^t W_s \, \text{sgn}(W_s) \, ds \right) =2\int_0^t \mathbb{E}(|W_s|) \, ds.$$

The integral on the right-hand side is strictly positive (in fact, it can be calculated explicitly, using the fact that $W_s \sim N(0,s)$ entails $\mathbb{E}(|W_s|) = \sqrt{(2s)/\pi}$). As $\mathbb{E}(X_t)=0$ this shows that $$\mathbb{E}(W_t^2 X_t) \neq \mathbb{E}(W_t^2) \mathbb{E}(X_t),$$ and therefore $W_t$ and $X_t$ are not independent.

  • $\begingroup$ Thanks! To be honest I can't see the point in looking at $W_t^2$. Please have a look at my own answer. Is it correct? $\endgroup$ – k.dkhk Jan 23 at 12:34

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